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%\author{王立庆（2019级数学与应用数学1班）}
\author{学号 \underline{\hspace{4cm}} 姓名  \underline{\hspace{4cm}} }
%\title{高等代数第六章：向量空间}
\title{第七章线性变换（7.1-7.3）考试解答 }
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\date{2023年4月11日}

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\begin{document}

\maketitle

%\begin{abstract}
%%主要内容：
%7.1. 
%7.2. 
%7.3. 

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%\end{abstract}

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\begin{enumerate}

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\item %1
设矩阵 $A=\begin{pmatrix} 1&1&2&0 \\ 2&4&5&6 \\ 4&6&9&6 \end{pmatrix}$. 
对 $\xi\in \mathbb{R}^4$, 定义 $\sigma(\xi)=A\xi\in \mathbb{R}^3$. 
求线性映射 $\sigma$ 的核空间和像空间。

\vspace{0.2cm}

{\color{red}解答：通过行初等变换，将矩阵 $A$ 化为行最简形，
$$ 
A=\begin{pmatrix} 1&1&2&0 \\ 2&4&5&6 \\ 4&6&9&6 \end{pmatrix}
\longrightarrow 
\begin{pmatrix} 1&0&3/2&-3 \\ 0&1&1/2&3 \\ 0&0&0&0 \end{pmatrix}.
$$

\begin{enumerate}

\item  求核空间。由 $A\xi=0$ 可得 
$$\xi = k_1\begin{pmatrix} -3/2 \\ -1/2 \\ 1 \\ 0 \end{pmatrix} + k_2\begin{pmatrix} 3 \\ -3 \\ 0 \\ 1 \end{pmatrix},
k_1,k_2\in\mathbb{R}. $$
因此核空间为 $\text{Ker}(\sigma)=L(\eta_1, \eta_2)$, 其中 
$$\eta_1 = \begin{pmatrix} -3 \\ -1 \\ 2 \\ 0 \end{pmatrix},
\eta_2=\begin{pmatrix} 3 \\ -3 \\ 0 \\ 1 \end{pmatrix}. $$

\item  求像空间。将矩阵 $A$ 按列向量分块，可以写成
$$ 
A=\begin{pmatrix} 1&1&2&0 \\ 2&4&5&6 \\ 4&6&9&6 \end{pmatrix}
= \begin{pmatrix} \alpha_1 & \alpha_2 & \alpha_3 & \alpha_4  \end{pmatrix}. 
$$
从矩阵 $A$ 的行最简形可得，向量组 $\{\alpha_1, \alpha_2\}$ 是向量组 $\{\alpha_1, \alpha_2, \alpha_3, \alpha_4\}$ 的一个极大线性无关组，而且 
\begin{align*}
\alpha_3 & = (3/2)\alpha_1+(1/2)\alpha_2, \\
\alpha_4 & = -3\alpha_1+3\alpha_2. 
\end{align*}
所以像空间为 $L(\alpha_1,\alpha_2)$. 

\end{enumerate}

}

\vspace{0.2cm}

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\item %2
设 $\sigma$ 是向量空间 $V$ 到自身的线性变换。
设核空间 $\text{Ker}(\sigma)=\{\theta\}$ 是零子空间。
证明 $\sigma$ 是单射。

\vspace{0.2cm}

{\color{red}解答：
\begin{enumerate}
\item[(1)]  设 $\sigma(\alpha)=\sigma(\beta)$. 
\item[(2)]  因为 $\sigma$ 保持线性运算，所以 $\sigma(\alpha-\beta)=\theta$, 所以 $\alpha-\beta\in \text{Ker}(\sigma)$. 
\item[(3)]  根据条件 $\text{Ker}(\sigma)=\{\theta\}$, 可得 $\alpha-\beta = \theta$, 即 $\alpha=\beta$. 
\item[(4)] 所以 $\sigma$ 是单射。 
\end{enumerate}

}

\vspace{0.2cm}

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\item %3
设线性变换 $\sigma\in L(V)$ 关于基 $\{\alpha_1,\alpha_2\}$ 的矩阵是
$A=\begin{pmatrix} 2&0 \\ 2&3 \end{pmatrix}$. 
(a) 求 $\sigma$ 关于基  
$ \left\{\begin{array}{rcl}
\beta_1 &=& 3\alpha_1 + 2\alpha_2 \\
\beta_2 &=& 4\alpha_1 + 3\alpha_2
\end{array} \right. $ 的矩阵。
(b) 设向量 $\xi=\alpha_1+2\alpha_2$. 求 $\sigma(\xi)$ 关于基 $\{\beta_1,\beta_2\}$ 的坐标。

\vspace{0.2cm}

{\color{red}解答：
\begin{enumerate} 

\item 第一问：

\begin{enumerate}
\item[(1)]  记 $\Phi = (\alpha_1,\alpha_2)$, $\Psi = (\beta_1,\beta_2)$, 这是向量空间 $V$ 的两个基。 

\item[(2)]  根据线性变换关于基的矩阵的定义，可得 
$\sigma(\Phi)=\Phi\cdot A$ 以及 $\sigma(\Psi) = \Psi\cdot B$. 

\item[(3)]  根据条件可得 $\Psi = \Phi\cdot P$, 其中 $P=\begin{pmatrix} 3&4 \\ 2&3 \end{pmatrix}$ 是从 $\Phi$ 到 $\Psi$ 的过渡矩阵。

\item[(4)]  因此所求矩阵为  
$$B=P{\,}^{-1}AP= 
\begin{pmatrix} 3&4 \\ 2&3 \end{pmatrix}^{-1}
\begin{pmatrix} 2&0 \\ 2&3 \end{pmatrix}
\begin{pmatrix} 3&4 \\ 2&3 \end{pmatrix}
=
\begin{pmatrix} -2&-12 \\ 2&9 \end{pmatrix}
\begin{pmatrix} 3&4 \\ 2&3 \end{pmatrix}
=\begin{pmatrix} -30&-44 \\ 24&35 \end{pmatrix}. 
$$

\end{enumerate}

\item  第二问：

\begin{enumerate}
\item[(1)]  记 $X=(1,2)^t$, 则 $\xi=\Phi\cdot X$. 

\item[(2)]  因为 $\sigma$ 是线性变换，所以 $\sigma(\xi) = \sigma(\Phi)\cdot X$. 

\item[(3)]  因为 $\sigma(\Phi)=\Phi\cdot A$, 所以 $\sigma(\xi) =\Phi\cdot A\cdot X$. 

\item[(4)]  因为 $\Psi = \Phi\cdot P$, 所以 $\Phi = \Psi\cdot P^{-1}$. 

\item[(5)]  由 (3)和(4)可得 $\sigma(\xi) = \Psi\cdot P^{-1}\cdot A\cdot X$. 

\item[(6)]  因此 $\sigma(\xi)$ 关于基 $\Psi$ 的坐标为 
$$
P^{-1}\cdot A\cdot X = 
\begin{pmatrix} 3&4 \\ 2&3 \end{pmatrix}^{-1}
\begin{pmatrix} 2&0 \\ 2&3 \end{pmatrix}
\begin{pmatrix} 1 \\ 2 \end{pmatrix}
=
\begin{pmatrix} -2&-12 \\ 2&9 \end{pmatrix}
\begin{pmatrix} 1 \\ 2 \end{pmatrix}
=
\begin{pmatrix} -26 \\ 20 \end{pmatrix}. 
$$

\end{enumerate}

\end{enumerate}

}

%\begin{lstlisting}[language=R]
%library(pracma)
%A=matrix(c(2,0,2,3),nrow=2,byrow=T)
%P=matrix(c(3,4,2,3),nrow=2,byrow=T)
%print(inv(P)%*%A%*%P)
%X=matrix(c(1,2),nrow=2,byrow=T)
%print(inv(P)%*%A%*%X)
%\end{lstlisting}

\vspace{0.2cm}

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%\item %4

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%{\color{red}解答：
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%
%}

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\end{enumerate}



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